# Ex 12.3, 2 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 12.3, 2 Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR. Given that Point P (3, 2, –4), Q (5, 4, –6) & R (9, 8, –10) are collinear Q must divide line segment PR in some ratio externally & internally . We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = mx2 + nx1m + n,my2+ my1m + n Let point Q (5, 4, –6) divide line segment P (3, 2, – 4), R (9, 8, – 10) in the ratio k : 1 Here, x1 = 3, y1 = 2, z1 = – 4 x2 = 9, y2 = 8, z2 = – 10 & m = k , n = 1 Putting values Q (5, 4, – 6) = k9+3k+1,k8+2k+1,k−10−4k+1 (5, 4, – 6) = 9𝑘 + 3k + 1,8𝑘 + 2k + 1,− 10𝑘 −4k+1 Comparing x – coordinate of Q 5 = 9k + 3k + 1 5 (k + 1) = 9k + 3 5k + 5 = 9k + 3 5k – 9k = 3 – 5 – 4k = – 2 k = −2−4 k = 12 So, k : 1 = 1 : 2 Thus, point Q divides PR in the Ratio 1 : 2

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.